Let us say, I have a distribution with some unknown parameter and I find the MLE of the parameter. I prove that the MLE is inadmissible using some other estimator. Why is this not in a contradiction to the general theorem on asymptotic optimality of maximum likelihood estimates?

  • $\begingroup$ what distribution are we talking about? can we know? $\endgroup$
    – carlo
    Oct 25, 2019 at 20:30
  • 3
    $\begingroup$ admissibility is not an asymptotic property. $\endgroup$ Oct 25, 2019 at 23:11
  • $\begingroup$ Please see stats.stackexchange.com/questions/189414/…. $\endgroup$
    – whuber
    Oct 26, 2019 at 14:07
  • 1
    $\begingroup$ In a track race, you can reach the finish line tied for first, yet everyone in the field could have been leading you at every step of the way. $\endgroup$
    – whuber
    Dec 20, 2022 at 19:08

1 Answer 1


Maximum likelihood estimators are the most efficient consistent estimators, yes. However, that is an asymptotic property. In a finite sample, all bets are off. Indeed, the Wikipedia article on maximum likelihood estimation mentions that MLEs have no optimum properties for finite samples.

You might wind up with a situation where, in a finite sample, a biased and inconsistent estimator has lower risk over the entire parameter space. That is, no matter the true value of the parameter, the biased and inconsistent estimator has lower risk than a maximum likelihood estimator.

The James-Stein estimator and Stein’s paradox might be of interest.


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